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MAVERICK [17]
3 years ago
12

6 (3x-1)-10x. with work please :)

Mathematics
1 answer:
STatiana [176]3 years ago
4 0
Hey there!
Let's break this expression into two parts:
6(3x-1) and -10x

To solve the first part, we need to use the distributive property which states:
a(b+c) = ab+ac

Applying that to this problem, we have:
6(3x) + 6(-1) =
18x - 6

Now, we can take that -10x and put it right back in:

18x - 6 - 10x

Combine like terms and subtract the 10x from the 18x to get:
8x - 6

Hope this helps!
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Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10
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The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

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Points are plotted at (-2, 2), (-2, -4), and (2, -4). A fourth point is drawn such that the four points can be connected to form
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Answer:24 square units

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Answer: She deposits 50 dollars per month right? But as known her parents deposited her 2,500. So, it will take her 50 months to complete the amount of  5000 dollars needed for college.

Step-by-step explanation: Im not good with equations but here  we go.

First: AS we know the account already had 2,500, and every month she deposits 50$.

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7 0
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